ucla stat 110b ivo dinov text: statistics for …dinov/courses_students.dir/03/spr...text:...

1

Stat 110B, UCLA, Ivo Dinov Slide 1

UCLA STAT 110BApplied Statistics for Engineering

and the Sciences

�Instructor: Ivo Dinov, Asst. Prof. In Statistics and Neurology

�Teaching Assistants: Brian Ng, UCLA Statistics

University of California, Los Angeles, Spring 2003http://www.stat.ucla.edu/~dinov/courses_students.html

Stat 110B, UCLA, Ivo DinovSlide 2

Course Organization

http://www.stat.ucla.edu/~dinov/courses_students.html

Software: No specific software is required. SYSTAT, R, SOCR resource, etc.

Text: Introduction to Probability and Statistics for Engineering and the Sciences 5th edition -- Jay Devore

Course Description, Class homepage,

online supplements,VOH’s, etc.


� Material Covered: (Devore, Chapters 7-14)

�Review of Key Concepts (ch 01-06)�Confidence Intervals (ch 07)�Single Sample Hypotheses testing (ch 08)� Inferences based on 2 samples (ch 09)�One- Two- and Three-Factor ANOVA (ch 10)�2k Factorial Designs (ch 11)�Linear Regression (ch 12)�Multiple & Nonlinear Regression (ch 13)�Goodness-of-Fit Testing (ch 14)

Course Organization


Overall Review

What is a statistic?- Any quantity whose value can be calculated from

sample data. It does not depend on any unknown parameter.

- Examples –

What are Random Variables?- A function from the sample space to the real number

line.Before any data is collected, we view all observations

and statistics as random variables


Properties of Expectation and Variance

� Let X be a random variable and a,b be constants. It follows that:

( )][)(

][][][

][][

][][

2

22

2

XVarXSDXEXEXVar

XVarabaXVarbXaEbaXE

=

−=

=+

+=+


Linear Combinations of Random Variables

Consider the collection of the independent random variables X1,…,Xn where E[Xi]=µi and Var[Xi]=σi

2, and let a1,…,an be constants. Define a random variable Y by

which is a linear combination of the Xi’s. It follows that

nnnnnn aaXEaXEaXaXaE µµ ++=++=++ ...][...][]...[ 111111

2221

21

21

2111

...

][...][]...[

nn

nnnn

aa

XVaraXVaraXaXaVar

σσ ++=

++=++

nn XaXaY ++= ...11

What if dependent?!?

2


Random Sample

X1,…,Xn are an IID random sample of size n if:

1. The Xi’s are independent random variables

2. Every Xi has the same (identical) probability distribution

These conditions are equivalent to the Xi’s being independent and identically distributed (iid) random variables


Sample Mean and Total of a Random Sample

The sample mean is given by the random variable defined as

The sample total is given by the random variable To defined as

�=

=n

iiX

nX

1

1

�=

=n

iio XT

1

X


Mean and Variance of To

For the total-sum random variable

T0 = X1+…+Xn

T0~N(nµ, nσ2 ).


Mean and Variance of X

For the total-sum random variable

= (1/n) (X1+…+Xn)

~N(µ, σ2/n ).

X

X


Let X1,…,Xn be a random sample from a normally distributed population with mean µand variance σ2, i.e. Xi~N(µ, σ2). It follows that the random variable Y = a1X1+…+anXn is normally distributed with mean a1µ,…,an µand variance a1

2σ2+…+an2 σ2. Hence, the

sample mean and the sample total of the random sample will be normally distributed.

Linear Combinations of Normal Random Variables from a Random Sample


The central limit theorem gives us information about the sample mean and the sample total for a “large” (n>30) random sample from a population that is not normally distributed. Specifically, it tells us that these will be approximately normally distributed. The larger n is, the better the approximation.

Central Limit TheoremArguably the most important theorem in Statistics (GUT theory)

3


When a certain type of electrical resistor is manufactured, the mean resistance is 4 ohms with a standard deviation of 1.5 ohms. If 36 batches are independently produced, what is the probability that the sample average resistance of the batch is between 3.5 and 4.5 ohms. What is the probability that the sample total resistance is greater than 140 ohms?

Do InteractiveNormalCurve & CLT_Sampling Distribution Applets from SOCR resource

Example – Central Limit Theorem


Uni- vs. Multi-modal histograms

� Number of clear humps on the frequency histogram plot determines the modality of a histogram plot.

0

5

10

15

20

25

30

35

0 40 80 130 More

02468

10

121416

2 30 58 86


Skewness & Symmetry of histograms

� A histogram is symmetric is the bars (bins) to the left of some point (mean) are approximately mirror images of those to the right of the mean.

� Histogram is skewed if it is not symmetric, the histogram is heavy to the left or right, or non-identical on both sides of the mean.

0

20

40

60

80

100

1 2 3 4 5 6 7 8 9 100

10

20

30

40

50

60

1 2 3 4 5 6 7 8 9 100

10

20

30

40

50

60

1 2 3 4 5 6 7

file:///C:/Ivo.dir/UCLA_Classes/Applets.dir/HistogramApplet.html


Skewness & Kurtosis

� What do we mean by symmetry and positive and negative skewness? Kurtosis? Properties?!?

� Skewness is linearly invariant Sk(aX+b)=Sk(X)

� Skewness is a measure of unsymmetry

� Kurtosis is (also linearly invariant) a measure of flatness

� Both are used to quantify departures from StdNormal

� Skewness(StdNorm)=0; Kurtosis(StdNorm)=3

( ) ( )4

1

4

31

3

)1(Kurtosis ;

)1(Skewness

SDN

YY

SDN

YYN

kk

N

kk

−

−=

−

−=

��==


Comparing 3 plots of the same data

Stem-and-leaf of strength N = 33Leaf Unit = 10

1 19 8 5 20 0334 5 20 10 21 00233 (8) 21 55668899 15 22 000111112 6 22 5 5 23 014 2 23 2 24 2 24 2 25 2 1 25 9

2000 2100 2200 2300 2400 2500 2600strength

2000 2100 2200 2300 2400 2500 2600strength

Figure 2.4.5 Three graphs of the breaking-strength data for}gear-teeth in positions 4 & 10 (Minitab output).


Important points

1. The distinction between a randomized experimentand an observational study is made at the time of result interpretation. The very same statistical analysis is carried for the two situations.

2. We’ve already stressed the importance of plotting data prior to stat-analysis. Plots have many important roles – prevent dangerous misconceptions from arising (data overlaps, clusters, outliers, skewness, trends in the data, etc.)

4


Analyzing Histogram Plots

� Modality – uni- vs. multi-modal (Why do we care?)

� Symmetry – how skewed is the histogram?

� Center of gravity for the Histogram plot – does it make sense?

� If center-of-gravity exists quantify the spread of the frequencies around this point.

� Strange patterns – gaps, atypical frequencies lying away from the center.


Measures of central tendency (location)

� Mean – sum of all observations divided by their number� Median – (second quartile, Q2) is the half-way-point for

the distribution, 50% of all data are greater than it and 50% are smaller than Q2.

� Mode – the (list of) most frequently occurring observation(s).

25% 25%25%

median25%

mean

Range [min : max]

mode


Measures of variability (deviation)

� Mean Absolute Deviation (MAD) –

� Variance –

� Standard Deviation –

�=

−−

=n

i i yyn

MAD11

1

( )�=

−−

==n

i i yyn

sVar1

22

11

( )�=

−−

===n

i i yyn

sVarSD1

2

11


Measures of variability (deviation)

� Example:� Mean Absolute Deviation–

� Variance –

� Standard Deviation –

� X={1, 2, 3, 4}.

�=

−−

=n

i i yyn

MAD11

1

( )�=

−−

==n

i i yyn

sVar1

22

11

( )�=

−−

=n

i i yyn

SD1

2

11

1 2 3 4m=2.5

MAD=4/3=1.33Var=5/3=1.67SD=1.3


Trimmed, Winsorized means and Resistancy

� A data-driven parameter estimate is said to be resistant if it does not greatly change in the presence of outliers.

� K-times trimmed mean

� Winsorized k-times mean:

�−

+=−=

kn

ki itk ykn

y1 )(2

1

[ ])(

1

2 )()1( )1()1(1kn

kn

ki ikwk ykyykn

y −

−−

+=+ ++++= �

Orderstatistic


Stationary or Non-Stationary Process?

�To assess stationarity:� Rigorous assessment: A stationary process has a constant

mean, variance, and autocorrelation through time/place.

� Visual assessment: (Plot the data – observed vs. time/place – the parameter we argue stationarity with respect to).

Time-Series Plot of the KWH Data

0

10

20

30

40

50

60

1 16 31 46 61 76 91 106

121

136

151

166

181

196

Date

KWH

Series1

5


Stationary or Non-Stationary Process?

� Visual assessment: (Plot the data – observed vs. time/place, etc., – parameter we argue stationarity with respect to).

Scatter Plot of the KWH Data

0

10

20

30

40

50

60

0 50 100 150 200 250

Date

KWH Series1


Moving Averages

� Signal, Noise, Filtering: Oftentimes high frequency oscillations in the data make it difficult to read/interpret the data.

117 33 49 65 81 97

113

129

145

161

177

193 S1

0

10

20

30

40

50

60

Series1


Moving Averages – next 10 values are averaged

� Signal, Noise, Filtering: Oftentimes high frequency oscillations in the data make it difficult to read/interpret the data.

Moving Average Effects on the Raw Data (KWH)

0102030405060

1 15 29 43 57 71 85 99 113

127

141

155

169

183

197

Date

KW

H

Raw KWH data Moving AverageStat 110B, UCLA, Ivo DinovSlide 28

Properties of probability distributions

� A sequence of number {p1, p2, p3, …, pn } is a probability distribution for a sample space S = {s1, s2, s3, …, sn}, if pr(sk) = pk, for each 1<=k<=n. The two essential properties of a probability distribution p1, p2, … , pn?

� How do we get the probability of an event from the probabilities of outcomes that make up that event?

� If all outcomes are distinct & equally likely, how do we calculate pr(A) ? If A = {a1, a2, a3, …, a9} and pr(a1)=pr(a2)=…=pr(a9 )=p;then

pr(A) = 9 x pr(a1) = 9p.

1 ;0 =≥ �k kk

pp


The conditional probability of A occurring given that B occurs is given by

pr(A | B) =pr(A and B)

pr(B)

Conditional Probability

Suppose we select one out of the 400 patients in the study and we want to find the probability that the cancer is on the extremitiesgiven that it is of type nodular: P = 73/125 = P(C. on Extremities | Nodular)

patientsnodular #sextremitieon cancer with patientsnodular #


pr(A and B) = pr(A | B)pr(B) = pr(B | A)pr(A)

Multiplication rule- what’s the percentage of Israelis that are poor and Arabic?

00.0728

0.14 1.0

All people in Israel

14% of these are Arabic

52% of this 14% are poor

7.28% of Israelis are both poor and Arabic(0.52 .014 = 0.0728)

Figure 4.6.1 Illustration of the multiplication rule.From Chance Encounters by C.J. Wild and G.A.F. Seber, © John Wiley & Sons, 2000.

6


Permutation: Number of ordered arrangements of robjects chosen from n distinctive objects

e.g. P63 = 6·5·4 =120.

Permutation & Combination

rr

rnn

nn PPP ·−=

)1()2)(1( +−…−−= rnnnnP rn


Combination: Number of non-orderedarrangements of r objects chosen from n distinctive objects:

Or use notation of e.g. 3!=6 , 5!=120 , 0!=1


!)!(!!/

rrnnrPC r

nrn −

==

( ) rn

nr C=

( ) 35!3!4

!773 ==


Combinatorial Identity:

Analytic proof: (expand both hand sides)Combinatorial argument: Given n object focus on one of

them (obj. 1). There are groups of size r that contain obj. 1 (since each group contains r-1 other elements out of n-1). Also, there are groups of size r, that do not contain obj1. But the total of all r-size groups of n-objects is !


( ) ( ) ( )111

−−− += n

rnr

nr

��

��

−−

11

rn

��

�� −

rn

1

��

��rn


Combinatorial Identity:

Analytic proof: (expand both hand sides)Combinatorial argument: Given n objects the number of

combinations of choosing any r of them is equivalent to choosing the remaining n-r of them (order-of-objs-not-important!)


( ) ( )nrn

nr −=


Examples

1. Suppose car plates are 7-digit, like AB1234. If all the letters can be used in the first 2 places, and all numbers can be used in the last 4, how many different plates can be made? How many plates are there with no repeating digits?

Solution: a) 26·26·10·10·10·10

b) P262 · P10

3 = 26·25·10·9·8·7


Examples

2. How many different letter arrangement can be made from the 11 letters of MISSISSIPPI?

(((( ))))(((( ))))(((( ))))(((( )))) (((( ))))(((( ))))(((( ))))(((( ))))11

54

94

112

22

64

104

111 ... ========

Solution: There are: 1 M, 4 I, 4 S, 2 P letters.Method 1: consider different permutations:

11!/(1!4!4!2!)=34650Method 2: consider combinations:

7


Examples

3. There are N telephones, and any 2 phones are connected by 1 line. Then how many lines are needed all together?

Solution: C2N = N (N - 1) / 2

If, N=5, complete graph with 5 nodes hasC2

5=10 edges.


Binomial theorem & multinomial theorem

Deriving from this, we can get such useful formula (a=b=1)

Also from (1+x)m+n=(1+x) m(1+x)n we obtain:

On the left is the coeff of 1kx(m+n-k). On the right is the same coeff in the product of (…+ coeff * x(m-i) +…) * (…+coeff * x(n-k+i) +…).

( ) knkn

k

nk

n baba −

=�=+ )(

0

( ) ( ) ( ) ( )nnnn

nn 112...10 +==+++

( ) ( )( )nik

k

i

mi

nmk −

=

+�=

0

Binomial theorem


Multinomial theorem

( ) knkn

k

nk

n baba −

=�=+ )(

0

( ) r

r

nr

nnnnnn

nr xxxxxx ... )...( 21

21 21

,...,,21 �=+++

Generalization: Divide n distinctive objects into r groups, with the size of every group n1 ,…,nr, and n1+n2+…+nr = n

( ) ( )( ) ( )!!...!

!...21

...

,...,,

1112121

r

nnnn

nnn

nn

nnnn nnn

nr

rr== −−−−−where


Sterling Formula for asymptotic behavior of n!

Sterling formula:

n

en

nn �

�

��

�×= π2!


Probability and Venn diagrams

Proposition

P(A1U A2U… U An)=

( )

( )( )��

��

�

iniin

nirji iriir

nii ii

n

i i

AAAP

AAAP

AAPAP

....)1(

.......)1(

...)(

211

...211 211

211 211

+

≤<<<≤

+

≤<≤=

−+

+−+

+−

�

��


Discrete Variables, Probabilities

8


Binomial Probabilities –the moment we all have been waiting for!

� Suppose X ~ Binomial(n, p), then the probability

� Where the binomial coefficients are defined by

nxppxn

xXP xnx ≤≤≤≤≤≤≤≤−−−−��

��

��======== −−−− 0 ,)1()( )(

nnnxxn

nxn

××××−−−−××××××××××××××××====−−−−

====��

��

�� )1(...321! ,! )!(

!

n-factorial


Prize ($) x 1 2 3Probability pr(x) 0.6 0.3 0.1

$ won

Total prize money = Sum; Average prize money = Sum/100 = 1 0.6 + 2 0.3 + 3 0.1 = 1.5

SumNumber of games won

What we would "expect" from 100 games add across row0.6 100 0.3 100 0.1 100

2 0.3 100 3 0.1 1001 0.6 100

Expected values

� The game of chance: cost to play:$1.50; Prices {$1, $2, $3}, probabilities of winning each price are {0.6, 0.3, 0.1}, respectively.

� Should we play the game? What are our chances of winning/loosing?

Theoretically Fair Game: price to play EQ the expected return!


)-1( = )sd( pnpX

For the Binomial distribution . . . mean

E(X) = n p,

X~Binomial(n, p) ��

X=Y1+Y2+Y3+..+Yn,where Yk ~Bernoulli(p),

E(Y1)=p ��E(X) = E(Y1+Y2+Y3+..+Yn)=np


Poisson Distribution – Definition

� Used to model counts – number of arrivals (k) on a given interval …

� The Poisson distribution is also sometimes referred to as the distribution of rare events. Examples of Poisson distributed variables are number of accidents per person, number of sweepstakes won per person, or the number of catastrophic defects found in a production process.


Functional Brain Imaging - Positron Emission Tomography (PET)

http://www.nucmed.buffalo.eduStat 110B, UCLA, Ivo DinovSlide 48

Poisson Distribution – Mean

� Used to model counts – number of arrivals (k) on a given interval …

� Y~Poisson( ), then P(Y=k) = , k = 0, 1, 2, …

� Mean of Y, µY = λ, since!

ek

k λλ −

λ

λλλλλλ

λλλ

λλλλ

λλλ

===−

=

=−

===

−∞

=

−∞

=

−−

∞

=

−∞

=

−∞

=

−

��

��

ee!

e)!1(

e

)!1(e

!e

!e)(

01

1

100

k

k

k

k

k

k

k

k

k

k

kk

kkk

kkYE

9


Poisson Distribution - Variance

� Y~Poisson( ), then P(Y=k) = , k = 0, 1, 2, …

� Variance of Y, σY = λ½, since

� For example, suppose that Y denotes the number of blocked shots (arrivals) in a randomly sampled gamefor the UCLA Bruins men's basketball team. Then a Poisson distribution with mean=4 may be used to model Y .

!ek

k λλ −λ

λλλσλ

==−== �∞

=

−

...!

e)()(0

22

k

k

Y kkYVar


Poisson as an approximation to Binomial

� Suppose we have a sequence of Binomial(n, pn)models, with lim(n pn) � λλλλ, as n�infinity.

� For each 0<=y<=n, if Yn~ Binomial(n, pn), then

� P(Yn=y)=�But this converges to:

� Thus, Binomial(n, pn) � Poisson(λλλλ)

ynn

y

n ppyn −−��

�� )1(

!)1(

yey

ppyn

npn

nyn

n

y

n

λλλ

− →−�

��

��

→×

∞→−

WHY?


Poisson as an approximation to Binomial

� Rule of thumb is that approximation is good if:

� n>=100� p<=0.01� λλλλ =n p <=20

� Then, Binomial(n, pn) � Poisson(λλλλ)


Example using Poisson approx to Binomial

� Suppose P(defective chip) = 0.0001=10-4. Find the probability that a lot of 25,000 chips has > 2 defective!

� Y~ Binomial(25,000, 0.0001), find P(Y>2). Note that Z~Poisson(λλλλ =n p =25,000 x 0.0001=2.5)

456.0!25.2

!15.2

!05.21

!5.21)2(1)2(

5.22

5.21

5.20

2

0

5.2

=��

��

� ++−

=−=≤−=>

−−−

=

−�

eee

ez

ZPZPz

z


Geometric, Hypergeometric, Negative Binomial

� X ~ Geometric(p), then the probability mass function is

Probability of first failure at xth trial.

� Ex: Stat dept purchases 40 light bulbs; 5 are defective.

Select 5 components at random.

Find: P(3rd bulb used is the first that does not work) = ?

21 1)( ;1)( ;)1()(

ppXVar

ppXEppxXP x −=−=−== −



� Hypergeometric – X~HyperGeom(x; N, n, M)Total objects: N. Successes: M. Sample-size: n (without

replacement). X = number of Successes in sample

Ex: 40 components in a lot; 3 components are defectives.

Select 5 components at random.

P(obtain one defective) = P(X=1) = ?

��

��

�

��

��

�

−−

��

��

�

==

nN

xnMN

xM

xXP )(

NMN

NMn

NnNXVar

NMnXE

−×××−−=

=

1)(

)(

10


Hypergeometric Distribution & Binomial

�Binomial approximation to Hyperheometric�

Ex: 4,000 out of 10,000 residents are against a new tax. 15 residents are selected at random.

PHyperGeom(at most 7 favor the new tax) = ? (0.78706)Demo: Applets.dir/ProbCalc.htm (PBin(Y<=7)=0.7869]HyperGeom(x; N=104, n=15, M=4x103) � Bin(x;n=15,p=0.4)

pNM

Nn ≈< then0.1),(usually small is

),;(/

),,;( pnxBinpNM

MnNxHyperGeomapproaches

≈�



� Negative binomial pmf [X ~ NegBin(r, p), if r=1 �Geometric (p)]

Number of trials until the rth success (negative, since number of successes (r) is fixed & number of trials (X) is random)

ppxXP x 1)1()( −−==

2)1()( ;)(

)1(11

)(

pprXVar

prXE

pprn

nXP rnr

−==

−��

��

�

−−

== − Find E(X) and Var(X)X=# of times one mustThrow a dice until theOutcome 1 occurs 4Times:X~NegBin(x;r=4,p=1/6)E(X)=24; Var(X)=120


Continuous RV’s

� A RV is continuous if it can take on any real value in a non-trivial interval (a ; b).

� PDF, probability density function, for a cont. RV, Y, is a non-negative function pY(y), for any real value y, such that for each interval (a; b), the probability that Y takes on a value in (a; b), P(a<Y<b) equals the area under pY(y) over the interval (a: b).

�

pY(y)

a b

P(a<Y<b)


Convergence of density histograms to the PDF

� For a continuous RV the density histograms converge to the PDF as the size of the bins goes to zero.

�


Measures of central tendency/variability for Continuous RVs

� Mean

� Variance

� SD

�∞

∞−

×= dyypy YY )(µ

�∞

∞−

×−= dyypy YYY )()( 22 µσ

�∞

∞−

×−= dyypy YYY )()( 2µσStat 110B, UCLA, Ivo DinovSlide 60

Facts about PDF’s of continuous RVs

� Non-negative

� Completeness

� Probability

yypY ∀≥ ,0)(

1)( =�∞

∞−

dyypY

� ×=<<b

aY dyypybYaP )()(

11


Continuous Distributions

� Uniform distribution

� Normal distribution

� Student’s T distribution

� F-distribution

� Chi-squared ( )

� Cauchy’s distribution

� Exponential distribution

� Poisson distribution, …

2χ


(Continuous) Uniform Distribution

f(x) =,αβ −

1βα <<x

0 , otherwise

2)( βα +=XE

12)()(

2αβ −=XVar,

Ex) Uniform, α = 2, β = 7

(a)

(b)

=≥ )4(XP=<< )5.53( XP

βα

αβ −1

x

f(x)

• X ~ Uniform Distribution with parameters α and β if

• random numbers follow Uniform between 0 and 1


(General) Normal Distribution

� Normal Distribution PDF: Y~Normal(µ, σµ, σµ, σµ, σ2222) ��

��∞−

−−

∞−

−−

==

∞<<∞−∀=

yx

y

YY

y

Y

dxedxxpyF

yeyp

2

2)(

2

2)(

2)()(

,2

)(

2

2

2

2

πσ

πσσµ

σµ


Continuous Distributions – Student’s T

� Student’s T distribution [approx. of Normal(0,1)]�Y1, Y2, …, YN IID from a Normal(µ;σ)�Variance σ2 is unknown

� In 1908, William Gosset (pseudonym Student) derived the exact sampling distribution of the following statistics

� T~Student(df=N-1), where Y

YYTσ

µˆ−=

( )1

ˆ 1

2

−

−=�

=

N

YYN

kk

Yσ


Density curves for Student’s t

∞∞∞∞

0 2 4- 2- 4

df = ×[i.e., Normal(0,1)]

df = 5df = 2

Figure 7.6.1 Student(df) density curves for various df.

We will come back to theT-distribution at the endof this chapter!


Continuous Distributions – χχχχ2 [Chi-Square]

�χ2 [Chi-Square] goodness of fit test:�Let {X1, X2,…, XN} are IID N(0, 1)�W = X1

2 + X22 + X3

2 + …+ XN2

�W ~ χ2(df=N)�Note: If {Y1, Y2, …, YN} are IID N(µ, σµ, σµ, σµ, σ2222), then

�And the Statistics W ~ χ2(df=N-1)�E(W)=N; Var(W)=2N

( )�=

−−

=N

kk YY

NYSD

1

221

1)(

)(1 22 YSDNW

σ−=

χ2

12


Continuous Distributions – F-distribution

� F-distribution is the ratio of two χχχχ2 random variables.

� Snedecor's F distribution is most commonly used in tests of variance (e.g., ANOVA). The ratio of two chi-squares divided by their respective degrees of freedom is said to follow an F distribution

( ) ( )��==

−−

=−−

=M

ll

N

kk XX

MXSDYY

NYSD

1

22

1

221

1)( ;1

1)(

);(1 );(1 22

22 XSDMWYSDNW

XX

YY

σσ−=−=

)1,1(F~)1/()1/(

21 −=−=−−= MdfNdf

MWNWF

XY

o


Continuous Distributions – Cauchy’s

� Cauchy’s distribution, X~Cauchy(t,s), t=location; s=scale� PDF(X):

� PDF(Std Cauchy’s(0,1)):

� The Cauchy distribution is (theoretically) important as an example of a pathological case. Cauchy distributions look similar to a normaldistribution. However, they have much heavier tails. When studying hypothesis tests that assume normality, seeing how the tests perform on data from a Cauchy distribution is a good indicator of how sensitive the tests are to heavy-tail departures from normality. The mean and standard deviation of the Cauchy distribution are undefined!!! The practical meaning of this is that collecting 1,000 data points gives no more accurate of an estimate of the mean and standard deviation than does a single point (Cauchy=Tdf=0�Tdf�Normal).

( ) (reals) x;)/)(1

1)(2

R∈−+

=stxs

xfπ

( )211)(

xsxf

+=

π


Continuous Distributions – Exponential

� Exponential distribution, X~Exponential(λ)� The exponential model, with only one unknown parameter, is the

simplest of all life distribution models.

� E(X)=1/ λ; Var(X)=1/ λ2; � Another name for the exponential mean is the Mean Time To Fail

or MTTF and we have MTTF = 1/ λ. � If X is the time between occurrences of rare events that happen on the average

with a rate l per unit of time, then X is distributed exponentially with parameter λ. Thus, the exponential distribution is frequently used to model the time interval between successive random events. Examples of variables distributed in this manner would be the gap length between cars crossing an intersection, life-times of electronic devices, or arrivals of customers at the check-out counter in a grocery store.

0 ;)( ≥= − xexf xλλ


Continuous Distributions – Exponential

� Exponential distribution, Example:

� On weeknight shifts between 6 pm and 10 pm, there are an average of 5.2 calls to the UCLA medical emergency number. Let X measure the time needed for the first call on such a shift. Find the probability that the first call arrives (a) between 6:15 and 6:45 (b) before 6:30. Also find the median time needed for the first call ( 34.578%; 72.865% ). �We must first determine the correct average of this exponential

distribution. If we consider the time interval to be 4x60=240 minutes, then on average there is a call every 240 / 5.2 (or 46.15) minutes. Then X ~ Exp(1/46), [E(X)=46] measures the time in minutes after 6:00 pm until the first call.

By-hand vs. ProbCalc.htm


Normal approximation to Binomial

� Suppose Y~Binomial(n, p)� Then Y=Y1+ Y2+ Y3+…+ Yn, where

� Yk~Bernoulli(p) , E(Yk)=p & Var(Yk)=p(1-p) ��

� E(Y)=np & Var(Y)=np(1-p), SD(Y)= (np(1-p))1/2

� Standardize Y:� Z=(Y-np) / (np(1-p))1/2

� By CLT �� Z ~ N(0, 1). So, Y ~ N [np, (np(1-p))1/2]

� Normal Approx to Binomial is reasonable when np >=10 & n(1-p)>10(p & (1-p) are NOT too small relative to n).


Normal approximation to Binomial – Example

� Roulette wheel investigation:� Compute P(Y>=58), where Y~Binomial(100, 0.47) –

�The proportion of the Binomial(100, 0.47) population having more than 58 reds (successes) out of 100 roulette spins (trials).

� Since np=47>=10 & n(1-p)=53>10 Normal approx is justified.

�Z=(Y-np)/Sqrt(np(1-p)) = 58 – 100*0.47)/Sqrt(100*0.47*0.53)=2.2

� P(Y>=58) �� P(Z>=2.2) = 0.0139� True P(Y>=58) = 0.177, using SOCR (demo!)� Binomial approx useful when no access to SOCR avail.

Roulette has 38 slots18red 18black 2 neutral

13


Normal approximation to Poisson

� Let X1~Poisson(λλλλ) & X2~Poisson(µµµµ) ��X1+ X2~Poisson(λ+µλ+µλ+µλ+µ)

� Let X1, X2, X3, …, Xk ~ Poisson(λλλλ), and independent,� Yk = X1 + X2 + ··· + Xk ~ Poisson(kλλλλ), E(Yk)=Var(Yk)=kλλλλ.

� The random variables in the sum on the right are independent and each has the Poisson distribution with parameter λλλλ.

� By CLT the distribution of the standardized variable (Yk − kλλλλ) / (kλλλλ)1/2 �� N(0, 1), as k increases to infinity.

� So, for kλλλλ >= 100, Zk = {(Yk − kλλλλ) / (kλλλλ)1/2 } ~ N(0,1).�� Yk ~ N(kλλλλ, (kλλλλ)1/2).


Normal approximation to Poisson – example

� Let X1~Poisson(λλλλ) & X2~Poisson(µµµµ) ��X1+ X2~Poisson(λ+µλ+µλ+µλ+µ)

� Let X1, X2, X3, …, X200 ~ Poisson(2222), and independent,� Yk = X1 + X2 + ··· + Xk ~ Poisson(400), E(Yk)=Var(Yk)=400.

� By CLT the distribution of the standardized variable (Yk − 400) / (400)1/2 �� N(0, 1), as k increases to infinity.

�Zk = (Yk − 400) / 20 ~ N(0,1)�� Yk ~ N(400, 400).�P(2 < Yk < 400) = (std’z 2 & 400) = �P( (2−400)/20 < Zk < (400−400)/20 ) = P( -20< Zk<0)

= 0.5


Poisson or Normal approximation to Binomial?

� Poisson Approximation (Binomial(n, pn) � Poisson(λλλλ) ):

�n>=100 & p<=0.01 & λλλλ =n p <=20� Normal Approximation

(Binomial(n, p) � N ( np, (np(1-p))1/2) )�np >=10 & n(1-p)>10

!)1(

yey

ppyn

npn

nyn

n

y

n

λλλ

− →−�

��

��

→×

∞→− WHY?


Areas under Standard Normal Curve – Example

� Many histograms are similar in shape to the standard normal curve. For example, persons height. The height of all incoming female army recruits is measured for custom training and assignment purposes (e.g., very tall people are inappropriate for constricted space positions, and very short people may be disadvantages in certain other situations). The mean height is computed to be 64 in and the standard deviation is 2 in. Only recruits shorter than 65.5 in will be trained for tank operation and recruits within ½ standard deviations of the mean will have no restrictions on duties.� What percentage of the incoming recruits will be trained to operate

armored combat vehicles (tanks)?

� About what percentage of the recruits will have no restrictions on training/duties?


Areas under Standard Normal Curve - Example � The mean height is 64 in and the standard deviation is 2 in.

� Only recruits shorter than 65.5 in will be trained for tank operation.What percentage of the incoming recruits will be trained to operate armored combat vehicles (tanks)?

� Recruits within ½ standard deviations of the mean will have no restrictions on duties. About what percentage of the recruits will have no restrictions on training/duties?

60 62 64 65.5 66 68

X �� (X-64)/265.5 �� (65.5-64)/2 = ¾Percentage is 77.34%

X �� (X-64)/265 �� (65-64)/2 = ½63 �� (63-64)/2 = -½

Percentage is 38.30%60 62 63 64 65 66 68


Gamma and Exponential Distributions

�Gamma Distribution�Gamma function :

Properties

- X ~ Gamma with parameters α and β if

for positive integer n

0>α 0>β0 , otherwise

=)(xf

βαα αΓβ

x

ex−

−1

)(1

0>x,

where ,

dxex x−∞

−�=Γ

0

1)( αα 0>α

)1()1()( −−= αΓααΓ)!1()( −= nnΓ

1)1( =Γ π=Γ )5.0(

14



�Exponential Distribution (cont’d)CDF : ββ

β

xxx

edxexXPxF−−

−==≤= � 11)()(0

βx

exFxXP−

=−=> )(1)(

-

Ex 1) X = response time at a certain on-line computer terminal

X ~ exponential with E(X) = 5(sec.).

(a)

(b)

, x > 0

, x > 0

=≤ )10( XP=≤≤ )105( XP

- Tail probability

P{X>x}β1

f(x)

xF(x)


#events in t: Poisson w. mean λt


Relationship to the Poisson Process# of events in any time interval t has a Poisson distribution w/parameter �� the distribution of the elapsed time between two successive events is exponential with parameter

tλλ

β 1=

Why? Poisson : P(no events in t) =

Let X = time until the first event.

Then P(no events in t) =

i.e., = CDF of exponential with or

tt

etetP λλ λλ −

−

==!0

)();0(0

tetXP λ−=> )(

tetXP λ−−=≤≤ 1)0(β

λ 1=λ

β 1=

Exponential w. 1/λ

tX


Lognormal Distribution

� X ~ lognormal with parameters µ and σ, if

� f (x) =

� E(X) = exp( µ + σ 2/ 2)

Var(X) = exp ( 2µ + σ 2) {exp (σ 2) –1}Ex) Let X ~ lognormal with parameter µ = 3.2 and σ, = 1

P( X > 8) =

),;(~)ln( σµxNX

2

2

2)(ln

21 σ

µ

σπ

−− x

ex

0≥x

0 , otherwise


Weibull Distribution

� X ~ Weibull Distribution with parameters α and β if

f (x) =

� If β = 1 ; (exponential with parameter )

� Useful in Reliability, life testing problems

βαβαβ xex −−1 , x > 0

0 , otherwise

xexf αα −=)(α1

)11()(1

βΓα β +=

−XE

})]11([)21({)( 22

βΓ

βΓα β +−+=

−XVar

.βαxexF −−=1)(


Beta Distribution

� Provides positive density only in an interval of finite length

X ~ Beta Distribution with parameters α and β if

11 )1()()(

)( −− −+ βα

βΓαΓβαΓ xx , 0 < x < 1 (α>0, β>0 )

0 , otherwise

βαα+

=)(XE)1()(

)(2 +++

=βαβα

αβXVar,

Ex)

X = proportion of TV sets requiring service during the first year

~ beta, α = 3 , β = 2 .

P(at least 80% of the model sold this year will require service in 1 year)

f (x) =


Marginal & Joint PDF’sCentral Limit Theorem (CLT)

15


Joint probability mass function

� The joint probability mass function of the discrete random variables X and Y, denoted as fXY(x,y) satisfies:

),(),()3(

1),()2(

0),()1(

yYxXPyxf

yxf

yxf

XY

xXY

y

XY

===

=

≥

��


Joint probability mass function – exampleThe joint density, P{X,Y}, of the number of minutes waiting to catch the first fish, X , and the number of minutes waiting to catch the second fish, Y, is given below.

P {X = i,Y = k } k 1 2 3

Row Sum P{ X = i }

1 i 2 3

0.01 0.02 0.08 0.01 0.02 0.08 0.07 0.08 0.63

0.11 0.11 0.78

Column Sum P {Y =k }

0.09 0.12 0.79 1.00

• The (joint) chance of waiting 3 minutes to catch the first fish and 3 minutes to catch the second fish is:

• The (marginal) chance of waiting 3 minutes to catch the first fish is: • The (marginal) chance of waiting 2 minutes to catch the first fish is (circle all

that are correct): • The chance of waiting at least two minutes to catch the first fish is (circle

none, one or more): • The chance of waiting at most two minutes to catch the first fish and at most

two minutes to catch the second fish is (circle none, one or more):


Marginal probability distributions

� Individual probability distribution of a random variable is referred to as its Marginal Probability Distribution.

� Marginal probability distribution of X can be determined from the joint probability distribution of X and other random variables.

� Example: Marginal probability distribution of X is found by summing the probabilities in each column,

for y, summation is done in each row.


Marginal probability distributions (Cont.)

� If X and Y are discrete random variables with joint probability mass function fXY(x,y), then the marginal probability mass function of X and Y are

where Rx denotes the set of all points in the range of (X, Y) for which X = x and Ry denotes the set of all points in the range of (X, Y) for which Y = y

�===xR

XYX YXfxXPxf ),()()(

�===Ry

XYY YXfyYPyf ),()()(


Mean and Variance

� If the marginal probability distribution of X has the probability function f(x), then

� R = Set of all points in the range of (X,Y).

� Example.

�� =��

�

��

�

�===

x RXY

x RXY

xXX

xx

yxxfyxfxxxfXE ),(),()()( µ

�=R

XY yxxf ),(

),()(),()(

),()()()()(

22

222

yxfxyxfx

yxfxxfxXV

XYR

Xx R

XYX

x RXYX

xXXX

x

x

��

� ��

−=−=

−=−==

µµ

µµσ


Central Limit Theorem:When sampling from almost any distribution,

is approximately Normally distributed in large samples.X

Central Limit Theorem – heuristic formulation

Show Sampling Distribution Simulation Applet:file:///C:/Ivo.dir/UCLA_Classes/Winter2002/AdditionalInstructorAids/SamplingDistributionApplet.html

16


Let be a sequence of independentobservations from one specific random process. Let and and and both be finite ( ). If , sample-avg,

Then has a distribution which approaches N(µ, σ2/n), as .

Central Limit Theorem –theoretical formulation

{{{{ }}}},...,...,X,XXk21

µµµµ====)(XE σσσσ====)(XSD∞∞∞∞<<<<∞∞∞∞<<<<<<<< || ;0 µµµµσσσσ ��

========

n

k kX

nnX

1

1

X∞∞∞∞→→→→n


The standard error of the mean

The standard error of the sample mean is an estimate of the SD of the sample mean

� i.e. a measure of the precision of the sample mean as an estimate of the population mean

�given by SE( )size Sampledeviation standard Sample =

ns

xS x =)E(

x

� Note similarity with

� SD( ) = . X n

σσσσ


TABLE 7.2.1 Cavendish's Determinations of the Mean Density of the Earth (g/cm3)

5.50 5.61 4.88 5.07 5.26 5.55 5.36 5.29 5.58 5.655.57 5.53 5.62 5.29 5.44 5.34 5.79 5.10 5.27 5.395.42 5.47 5.63 5.34 5.46 5.30 5.75 5.68 5.85

So urce : C avendis h [1798].

Cavendish’s 1798 data on mean density of the Earth, g/cm3, relative to that of H2O

Sample mean

and sample SD =

Then the standard error for these data is:

3/ 447931.5 cmgx ====

3/ 2209457.0 cmgX

S ====

04102858.029

2209457.0)( ============n

SXSE X


� For random samples from a Normal distribution,

is exactly distributed as Student(df = n - 1)� but methods we shall base upon this distribution for T work

well even for small samples sampled from distributions which are quite non-Normal.

� df is number of observations –1, degrees of freedom.

)()(

XSEXT µµµµ−−−−====

Student’s t-distribution

Recall that for samples from N( µ , σ )

)1,0(~/

)()()( N

nX

XSDXZ

σσσσµµµµµµµµ −−−−====−−−−====

Approx/ExactDistributions


Inference & Estimation


Parameters, Estimators, Estimates …

�E.g., We are interested in the population mean diameter (parameter) of washers the sample-average formula represents an estimator we can use, where as the value of the sample averagefor a particular dataset is the estimate (for the mean parameter).

{ }( )

( )1900.01913.01896.032 .1903.0y

1900.01913.01896.031yestimate

0.1900 0.1913, 0.1896, :Data

1 estimator ;parameter

about How

1

++==++==

==== �

=

y

YY

NY

N

kkYµ

17


A 95% confidence interval

� A type of interval that contains the true value of a parameter for 95% of samples taken is called a 95%confidence interval for that parameter, the ends of the CI are called confidence limits.

� (For the situations we deal with) a confidence interval (CI) for the true value of a parameter is given by

estimate t standard errors (SE)±

TABLE 8.1.1 Value of the Multiplier, t , for a 95% CI

df : 7 8 9 10 11 12 13 14 15 16 17t : 2.365 2.306 2.262 2.228 2.201 2.179 2.160 2.145 2.131 2.120 2.110

df : 18 19 20 25 30 35 40 45 50 60 t : 2.101 2.093 2.086 2.060 2.042 2.030 2.021 2.014 2.009 2.000 1.960

∞


(General) Confidence Interval (CI)

� A level L confidence interval for a parameter (θ), is an interval (θ1^ , θ2^), where θ1^ & θ2^, are estimators of θ, such that P(θθθθ1111^ < θ <θ <θ <θ < θθθθ2222^) = L.

� E.g., C+E model: Y = µ+ε. Where ε ∼ Ν(0, σε ∼ Ν(0, σε ∼ Ν(0, σε ∼ Ν(0, σ 2222)))), then by CLT we have Y_bar ~ Ν(µ, σΝ(µ, σΝ(µ, σΝ(µ, σ2222/n)

� n½(Y_bar - µµµµ)/σ σ σ σ ~ Ν(0, σΝ(0, σΝ(0, σΝ(0, σ2222)))).� L = P ( z(1-L)/2 < n½(Y_bar - µµµµ)/σ σ σ σ < z(1+L)/2 ),

where zq is the qth quartile.

� E.g., 0.95 = P ( z0.025 < n½(Y_bar - µµµµ)/σσσσ < z0.975 ),

Area=?


Most of

the table

24.83

o

24.83

500th

100th

10th9th8th7th6th5th4th3rd2nd1st

1000th999th998th997th996th995th994th993rd992nd991st

502nd501st

96.0%

94.0%

90.0%88.9%100%100%100%100%100%100%100%100%

95.2%95.2%95.2%95.2%95.2%95.2%95.2%95.2%95.2%95.2%

96.0%96.0%

..........

..........

.......... ..........

..........

..........

SampleCoverage

to date

True mean

True mean

24.8424.82

Figure 8.1.2 Samples of size 10 from a Normal(µ=24.83, s=.005) distribution and their 95% confidence intervals for µ..

From Chance Encounters by C.J. Wild and G.A.F. Seber, © John Wiley & Sons, 1999.

How many of the previous samples contained the true mean?


Confidence Interval for the true (population) mean µ:sample mean t standard errors

or

±

1 and )SE( where),se( −==± ndfn

sxxtx x

CI for population mean

TABLE 8.1.1 Value of the Multiplier, t , for a 95% CI

df : 7 8 9 10 11 12 13 14 15 16 17t : 2.365 2.306 2.262 2.228 2.201 2.179 2.160 2.145 2.131 2.120 2.110

df : 18 19 20 25 30 35 40 45 50 60 t : 2.101 2.093 2.086 2.060 2.042 2.030 2.021 2.014 2.009 2.000 1.960

∞


Effect of increasing the confidence level

80% CI, x ± 1.282 se(x)

90% CI, x ± 1.645 se(x)

95% CI, x ± 1.960 se(x)

99% CI, x ± 2.576 se(x)

Figure 8.1.3 The greater the confidence level, the wider the interval

From Chance Encounters by C.J. Wild and G.A.F. Seber, © John Wiley & Sons, 2000.

80% CI, x ± 1.282 se(x)

90% CI, x ± 1.645 se(x)

99% CI, x ± 2.576 se(x)

95% CI, x ± 1.960 se(x)

ConfidenceLevel

Increase

Increases the size

of the CI

Why?


Effect of increasing the sample size

Passage time

n = 90 data points

n = 40 data points

n = 10 data points

24.83 24.8424.82

Figure 8.1.4 Three random samples from a Normal(µ=24.83, s =.005) distribution and their 95% confidence intervals for µ.

From Chance Encounters by C.J. Wild and G.A.F. Seber, © John Wiley & Sons, 2000,

To double the precision we need four times as many observations.

IncreaseSample

Size

Decreases the size

of the CI

18


Confidence intervals – non-symmetric case

� A marine biologist wishes to use male angelfish for an experiment and hopes their weights don't vary much. In fact, a previous random sample of n = 16 angelfish yielded the data below

� {y1; … ; yn} = { 5.1; 2.5; 2.8; 3.4; 6.3; 3.6; 3.9; 3.0; 2.7; 5.7; 3.5; 3.6; 5.3; 5.1; 3.5; 3.3}

� Sample statistics from these data include Avg. = 3.96 lbs, s2 = 1.35 lbs, n = 16.

� Problem: Obtain a 100(1- α)% CI(σ2).� Point Estimator for σ2? How about sample variance, s2?� Sampling theory for s2? Not in general, but under Normal

assumptions ...� If a random sample {Y1; …;Yn} is taken from a normal population

with mean µ and variance σ2,then standardizing, we get a sum of squared N(0,1)



� {y1; … ; yn} = { 5.1; 2.5; 2.8; 3.4; 6.3; 3.6; 3.9; 3.0; 2.7; 5.7; 3.5; 3.6; 5.3; 5.1; 3.5; 3.3}

� Problem: Obtain a 100(1- α)% CI(σ2).� If a random sample {Y1; …;Yn} is taken from a normal

population with mean µ and variance σ2,then standardizing, we get a sum of squared N(0,1) ( )

212

12

2

2~)1(

−==� −

=−ndf

nk k YYSn χ

σσ

( )

��

�

�

��

�

�

≤−

≤=−��

��

� −=

��

��

� −−

� 2

2 ,12

12

2

21 ,1

1 αα χχσ

αn

nk k

n

YYP

For a=0.05, say. Need:100(1- αααα)% CI(σσσσ2).



� {y1; … ; yn} = { 5.1; 2.5; 2.8; 3.4; 6.3; 3.6; 3.9; 3.0; 2.7; 5.7; 3.5; 3.6; 5.3; 5.1; 3.5; 3.3}

� Problem: Obtain a 100(1- α)% CI(σ2).

� χ2(15; 0.025)=27:49 and χ2(15; 0.975)=6:26 �� This yields the CI, the sample variance is s2=1.35. Note

the CI is NOT symmetric (0.74 ; 3.24)

( ) ( )2

21 ,1

12

2

2 ,1

12

2

��

�� −−

� = −

��

�� −

� = −≤≤

αχαχσ

n

nk YkY

n

nk YkY


Prediction vs. Confidence intervals

� Confidence Intervals (for the population mean µµµµ):

� Prediction Intervals: L-level prediction interval (PI) for a new value of the process Y is defined by:

( )

.mean processunknown theofestimator an as obtained

is ,ˆ valuepredicted thewhereL)/2(1 1,-nL)/2(1 1,-n tˆ ; tˆ ˆˆ

µ

σσYnewY

newnew YY=

++ ×+×−

��

��

� ×+

× ++

nt

Y ; n

t– Y L)/2(1 1,-nL)/2(1 1,-n ˆˆ σσ


Prediction vs. Confidence intervals – Differences?

� Confidence Intervals (for the population mean µµµµ):

� Prediction Intervals:( )

( )n

n

ky

ky

nY newnew

newnewnew

Y

YYYY11

12

11)(ˆˆ

ˆˆˆ

ˆ where;tˆ ; tˆL)/2(1 1,-nL)/2(1 1,-n

+×�

=−

−== −

=×+×− ++

σσ

σσ

( )�

=−

−==

��

��

� ×+

× ++

n

ky

ky

nY

12

11)(ˆˆ

ˆˆn

t Y ;

nt

– Y L)/2(1 1,-nL)/2(1 1,-n

σσ

σσ

Which SDis bigger?!?


Significance Testing –Using Data to Test Hypotheses

19


Example – Carbon content in Steel

Percentage of C (Carbon) in 2 random samples taken from 2 steel shipments are measured and summarized below. The question is to determine if there are statistically significant differences between the shipments.

0.0823.1882

0.0863.62101

S2Y_N#


Measuring the distance between the true-value and the estimate in terms of the SE’s

�Intuitive criterion: Estimate is credible if it’s not far-away from its hypothesized true-value!

�But how far is far-away?�Compute the distance in standard-terms:

�Reason is that the distribution of T is known in some cases (Student’s t, or N(0,1)).

�The estimator (obs-value) is typical/atypical if it is close to the center/tail of the distribution.

SEterValueTrueParameEstimatorT −−−−====


Comparing CI’s and significance tests

� These are different methods for coping with the uncertainty about the true value of a parameter caused by the sampling variation in estimates.

� Confidence interval: A fixed level of confidence is chosen. We determine a range of possible values for the parameter that are consistent with the data (at the chosen confidence level).

� Significance test: Only one possible value for the parameter, called the hypothesized value, is tested against the data. We determine the strength of the evidence (confidence) provided by the data against the proposition that the hypothesized value is the true value.


Review

�Are the carbon contents in the two steel shipments any different?

0.0823.18820.0863.62101

S2Y_N#

0.025%

025.0,7 == αdft12.3

8082.0

10086.0

44.0)2ˆ1ˆ(

18.362.3

0- Est_2-Est_10t

=+

=

=−

−=

==

µµSE

SE


Guiding principles

We cannot rule in a hypothesized value for a parameter, we can only determine whether there is evidence, provided by the data, to rule out a hypothesized value.

The null hypothesis tested is typically a skeptical reaction to a research hypothesis

Hypotheses


STEP 1 Calculate the test statistic ,

estimate - hypothesized valuestandard error

[This te lls us ho w many s tandard e rro rs the es tima te is abo ve the hypo thes ized

va lue (t0 po s itive ) o r be lo w the hypo thes ized va lue (t0 nega tive).]

STEP 2 Calculate the P -value using the following table.

STEP 3 Interpret the P -value in the context of the data.

Using to test H 0: θθθθ = θθθθ 0 versus some alternative H 1.ˆ θ

t0 =ˆ θ −θ0

se( ˆ θ )=

The t-test

20


Alternative Evidence against H0: θθθθ > θθθθ 0

hypothesis provided by P -value

H 1: θ > θ0 too much bigger than θ0 P = pr(T t 0)(i.e., - θ0 too large)

H 1: θ < θ0 too much smaller than θ0 P = pr(T t 0)(i.e., - θ0 too negative)

H 1: θ θ0 too far from θ0 P = 2 pr(T |t 0|)(i.e., | - θ0| too large)

where T ~ Student(df )

≠

ˆ θ

ˆ θ

ˆ θ

≤

≥

≥ˆ θ

ˆ θ

ˆ θ

The t-test


Interpretation of the p-value

TABLE 9.3.2 Interpreting the S ize of a P -Value

Translation> 0.12 (12%) No evidence against H0

0.10 (10%) Weak evidence against H0

0.05 (5%) Some evidence against H0

0.01 (1%) Strong evidence against H0

0.001 (0.1%) Very Strong evidence against H0

Approximate sizeof P -Value


Is a second child gender influenced by the gender of the first child, in families with >1 kid?

� Research hypothesis needs to be formulated first before collecting/looking/interpreting the data that will be used to address it. Mothers whose 1st child is a girl are more likely to have a girl, as a second child, compared to mothers with boys as 1st children.

� Data: 20 yrs of birth records of 1 Hospital in Auckland, NZ.

Male Female Total Male 3,202 2,776 5,978Female 2,620 2,792 5,412Total 5,822 5,568 11,3901st

Chi

ld

Second Child Gender


Analysis of the birth-gender data

� Samples are large enough to use Normal-approx. Since the two proportions come from totally diff. mothers they are independent � use formula 8.5.5.a

8109.1)0

tPr(

2

)2

ˆ1(2

ˆ

1

)1

ˆ1(1

ˆ2

ˆ1

ˆ

2ˆ

1ˆ

02

ˆ1

ˆ

49986.5edValueHypothesiz-Estimate0

t

−−−−××××====≥≥≥≥====−−−−

====−−−−

++++−−−−

−−−−====

��

�� −−−−

−−−−−−−−

============

TvalueP

n

pp

n

pp

pp

ppSE

ppSE


Analysis of the birth-gender data

� We have strong evidence to reject the H0, and hence conclude mothers with first child a girl a more likely to have a girl as a second child.

� Practical vs. Statistical significance:� How much more likely? A 95% CI:

CI (p1- p2) =[0.033; 0.070]. And computed by:

%]7; %3[0093677.096.10515.02

)2

ˆ1(2

ˆ

1

)1

ˆ1(1

ˆ96.1

2ˆ

1ˆ

2ˆ

1ˆ96.1

2ˆ

1ˆSEestimate

====××××±±±±

====−−−−

++++−−−−

××××±±±±−−−−

====��

��

�� −−−−××××±±±±−−−−====××××±±±±

n

pp

n

pppp

ppSEppz


Relation Among Various Continuous Distributions

Normal (X)2,σµ

Normal (Z)1,0

σµ−= XZ

Lognormal (Y)2,σµ

YX ln= XeY =Chi-square ( )

n

2χ

� == n

i iZ1

2χ

Gammaβα ,

2,2/ == βα n

Exponential(X)β

1=α

n=2

Weibullβγ,

1=γ

Uniform(U)1,0

UX lnβ−=

Uniform(X)βα ,

αβα

−−= XU ααβ +−= UX )(

Betaβα , 1== βα

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